Why do we assume that the residue field extension is separable in the definition of unramified

Question

Neukrich defines a finite extension $L / K$ of henselian fields to be unramified if the extension of residue fields $\lambda / \kappa$ is separable and $[L : K] = [\lambda : \kappa]$ (emphasis added).

I see how the condition that the residue field extension be separable is necessary for various nice theorems to hold for such extensions, but I don't have much intuition for why this condition is necessary, desirable, or natural.

Is there a way of seeing why we would want to require the residue field extension to be separable before we see that this condition is necessary for everything to work out how we want it to? I suspect perhaps having an inseparable residue field extension corresponds to something unpleasant geometrically, but I'm not well-enough versed in algebraic geometry to see what that might be.

Answer 1

You have no intuition for this situation because you never had to work with inseparable residue field extensions before, so you don't know what to expect about the role of that.

What we are facing here is not really an issue about henselian fields: it shows up when you want to define the notion of unramified primes in an extension of Dedekind domains. I have in mind here not just rings of integers of number fields, but general Dedekind domains. Unramified, whatever it ultimately means, should turn out to mean "not ramified", so the task of deciding when a prime is unramified is the same as deciding when it is ramified: the concepts should be complementary to each other.

For number fields, ramified primes downstairs are the prime factors of the discriminant. For a number field $K$, a prime $p$ ramifies in $\mathcal O_K$ if and only if $p \mid {\rm disc}(\mathcal O_K)$. More generally, if $E/F$ is an extension of number fields, then a nonzero prime ideal $\mathfrak p$ in $\mathcal O_F$ ramifies in $\mathcal O_E$ if and only if $\mathfrak p \mid \mathfrak d_{E/F}$, where $\mathfrak d_{E/F}$ is the discriminant ideal of $E/F$ (or more accurately $\mathcal O_E/\mathcal O_F$).

For an arbitrary Dedekind domain $A$, its integral closure $B$ in a finite extension of the fraction field of $A$ is also a Dedekind domain (yes, we don't need to assume the finite extension of the fraction field is separable: $B$ is Dedekind no matter what, but most books don't cover the most general case). There is a generalization to the ring extension $B/A$ of the discriminant ideal construction from number fields. It's a nonzero ideal $\mathfrak d_{B/A}$ in $A$ and it has a surprise: the prime ideals dividing $\mathfrak d_{B/A}$ are exactly the nonzero prime ideals $\mathfrak p$ in $A$ such that for some prime $\mathfrak P$ dividing $\mathfrak pB$, either (i) $\mathfrak P$ has multiplicity greater than $1$ as a factor of $\mathfrak pB$ or (ii) the residue field extension $B/\mathfrak P$ is inseparable over $A/\mathfrak p$. So these two conditions show up together!

We want the rule "ramified primes are the prime factors of the discriminant" to continue to be true for the Dedekind domain extension $B/A$, so we are forced to define "ramified" to mean that either (i) or (ii) occurs: some multiplicity is bigger than $1$ or some residue field extension is inseparable. Negating, we are forced to define "unramified" to mean neither (i) nor (ii) occurs: all multiplicities are $1$ and all residue field extensions are separable.

Dino Lorenzini addresses this issue in his book An Invitation to Arithmetic Geometry. On p. 101, he writes

Our reader may (or should!) be puzzled by the fact that we wish to give a name (namely, "ramified") to the maximal ideals $M$ of $B$ whose ramification index is bigger than one or whose associated residue field extension is not separable. [...] it is clear that we should be interested in understanding the integers $e_{M/P}$ for all $M \in {\rm Max}(B)$. But there is no reason, a priori, to be interested in the type (separable or inseparable) of the residue field extension associated to a maximal ideal of $B$.[...] Indeed, it is only after having tried to characterize the set of maximal ideals of $B$ whose ramification index is bigger than one that one realizes that there is no easy characterization for this set, and that it is the set of maximal ideals ramified over $A$ (as in the above definition) that lends itself to a beautiful description.

I suggest you take a look at Lorenzini's book.

In Borevich and Shafarevich's Number Theory, they work with Krull rings (which they call "rings with a theory of divisors"). These are a generalization of Dedekind domains, and they define ramified primes to be those with a repeated prime factor upstairs (see p. 203), and unramified otherwise. That's the more intuitive definition you want, but after the discriminant is constructed in exercise 17 of Section 5 of Chapter 3, exercise 18 asks the reader to show a prime divides the discriminant if and only if it ramified in their sense or has an inseparable residue field extension. So you can make up whatever terminology you want, but a choice of words can't prevent the two conditions "some mult. $> 1$" and "some res. field extension is inseparable" from being inextricably linked.

Answer 2

If you have a finite separable extension $K\subset F$ of degree $d$, then $F\otimes_K \overline{K}\cong \overline{K}^d$ as rings. On the level of schemes, this corresponds to $d$ distinct points.

If you have a finite inseparable extension, $K\subset F$, then $F\otimes_K \overline{K}$ is not isomorphic to $\overline{K}^d$ as a ring. It is isomorphic to a product of fewer than $d$ artinian local rings, all of which have some nilpotents - this means on the level of schemes, we have fewer than $d$ distinct points and some of them are non-reduced, which should be thought of as some points "coming together".

This "coming together" is the same thing that happens when we talk about ramification in geometry: a ramification point is one where you have fewer points in the fiber than you should (see the picture in the link). One instructive example is the projection from $y=x^2$ on to the $y$-axis - for $a>0$, the pre-image of $(0,a)$ is two points, $(\pm\sqrt{a},a)$, while for $a=0$, it's one point $(0,0)$.